Question

Three resistors of 3 Ω, 1.5 Ω and 1 Ω are connected in parallel. The combination of the above resistors is connected in series to a resistance of 8.5 Ω and then to a battery with an e.m.f. 4.5 V. If the internal resistance of the cell is negligible:

1. Draw the circuit diagram.
2. Calculate the current in the main circuit.
3. Calculate the current in each of the resistors in parallel.

Answer 1

(a)

(b) Equivalent resistance of 3 Ω, 1 Ω and 1.5 Ω in parallel

\[ \frac{1}{R_{p}}=\frac{1}{3}+\frac{1}{1.5}+\frac{1}{1}=\frac{2}{1} \text { or } \mathrm{R}_{\mathrm{p}}=\frac{1}{2} \Omega=0.5 \Omega \]

∴ Total equivalent resistance in series \(=0.5+8.5=9 \Omega\)

\[ \therefore \text{Current in main circuit, } \mathrm{I} = \frac{V}{R} = \frac{4.5}{9} = 0.5 \ \text{A}\]

(c) P.D. across parallel circuit, \(\mathrm{V}=\mathrm{I} . \mathrm{R}=0.5 \times 0.5=0.25 \text{~V}\)

\[ \text { ∴ Current in } 3 \Omega \text { resistor, } I_{1}=\frac{V}{R}=\frac{0.25}{3}=0.083 \ \text{A} \]

\[ \text{Similarly, current in } 1.5 \ \Omega \text{ resistor, } I_{2} = \frac{V}{R} = \frac{0.25}{1.5} = 0.167 \ \text{A}\]

\[ \text { Current in } 1 \Omega \text { resistor, } I_{3}=\frac{V}{R}=\frac{0.25}{1}=0.250\ \text {A} \]